Solve `sin 2x+cos 4x=2`.

Answer

To solve the equation \( \sin 2x + \cos 4x = 2 \), we will follow these steps:

### Step 1: Analyze the equation
The maximum value of \( \sin 2x \) is 1 and the maximum value of \( \cos 4x \) is also 1. Therefore, the maximum value of \( \sin 2x + \cos 4x \) is \( 1 + 1 = 2 \). This means that the equation can only hold true if both \( \sin 2x \) and \( \cos 4x \) are at their maximum values simultaneously.

### Step 2: Set conditions for maximum values
To achieve the maximum value of 2, we need:
\[ \sin 2x = 1 \quad \text{and} \quad \cos 4x = 1 \]

### Step 3: Solve for \( \sin 2x = 1 \)
The equation \( \sin 2x = 1 \) occurs at:
\[ 2x = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus,
\[ x = \frac{\pi}{4} + n\pi \]

### Step 4: Solve for \( \cos 4x = 1 \)
The equation \( \cos 4x = 1 \) occurs at:
\[ 4x = 2k\pi \quad \text{for } k \in \mathbb{Z} \]
Thus,
\[ x = \frac{k\pi}{2} \]

### Step 5: Find common solutions
Now we need to find values of \( x \) that satisfy both conditions:
1. \( x = \frac{\pi}{4} + n\pi \)
2. \( x = \frac{k\pi}{2} \)

To find common solutions, we can set:
\[ \frac{\pi}{4} + n\pi = \frac{k\pi}{2} \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \frac{k\pi}{2} - n\pi = \frac{\pi}{4} \]
Multiplying through by 4 to eliminate the fractions:
\[ 2k\pi - 4n\pi = \pi \]
This simplifies to:
\[ 2k - 4n = 1 \]

### Step 7: Solve for integers \( k \) and \( n \)
This equation implies that \( 2k = 4n + 1 \), which means \( 4n + 1 \) must be even. However, since \( 4n \) is always even, \( 1 \) makes the left side odd. Therefore, there are no integer solutions for \( k \) and \( n \).

### Conclusion
Since there are no integer solutions for \( k \) and \( n \), the original equation \( \sin 2x + \cos 4x = 2 \) has no solutions.

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